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1, \(x\left(x+\dfrac{2}{3}\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x+\dfrac{2}{3}=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{-2}{3}\end{matrix}\right.\)
2, a, \(\left|x+\dfrac{4}{6}\right|\ge0\)
Để \(\left|x+\dfrac{4}{6}\right|\) đạt GTNN thì \(\left|x+\dfrac{4}{6}\right|=0\)
\(\Leftrightarrow x+\dfrac{4}{6}=0\Rightarrow x=\dfrac{-2}{3}\)
Vậy, ...
b, \(\left|x-\dfrac{1}{3}\right|\ge0\)
Để \(\left|x-\dfrac{1}{3}\right|\) đạt GTLN thì \(\left|x-\dfrac{1}{3}\right|=0\)
\(\Leftrightarrow x-\dfrac{1}{3}=0\Rightarrow x=\dfrac{1}{3}\)
Vậy, ...
1)
a)
\(x\cdot\left(x+\dfrac{2}{3}\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x+\dfrac{2}{3}=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{2}{3}\end{matrix}\right.\)
2)
a)
\(\left|x+\dfrac{4}{6}\right|\ge0\)
Dấu \("="\) xảy ra khi \(x+\dfrac{4}{6}=0\Leftrightarrow x=\dfrac{-4}{6}\Leftrightarrow x=\dfrac{-2}{3}\)
Vậy \(Min_{\left|x+\dfrac{4}{6}\right|}=0\text{ khi }x=\dfrac{-2}{3}\)
b)
\(\left|x-\dfrac{1}{3}\right|\ge0\)
Dấu \("="\) xảy ra khi \(x-\dfrac{1}{3}=0\Leftrightarrow x=\dfrac{1}{3}\)
Vậy \(Min_{\left|x-\dfrac{1}{3}\right|}=0\text{ khi }x=\dfrac{1}{3}\)
Bài 1:
a)
\(|x+\frac{4}{15}|-|-3,75|=-|-2,15|\)
\(\Leftrightarrow |x+\frac{4}{15}|-3,75=-2,15\)
\(\Leftrightarrow |x+\frac{4}{15}|=-2,15+3,75=\frac{8}{5}\)
\(\Rightarrow \left[\begin{matrix} x+\frac{4}{15}=\frac{8}{5}\\ x+\frac{4}{15}=-\frac{8}{5}\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=\frac{4}{3}\\ x=\frac{-28}{15}\end{matrix}\right.\)
b )
\(|\frac{5}{3}x|=|-\frac{1}{6}|=\frac{1}{6}\)
\(\Rightarrow \left[\begin{matrix} \frac{5}{3}x=\frac{1}{6}\\ \frac{5}{3}x=-\frac{1}{6}\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=\frac{1}{10}\\ x=-\frac{1}{10}\end{matrix}\right.\)
c)
\(|\frac{3}{4}x-\frac{3}{4}|-\frac{3}{4}=|-\frac{3}{4}|=\frac{3}{4}\)
\(\Leftrightarrow |\frac{3}{4}x-\frac{3}{4}|=\frac{3}{2}\)
\(\Rightarrow \left[\begin{matrix} \frac{3}{4}x-\frac{3}{4}=\frac{3}{2}\\ \frac{3}{4}x-\frac{3}{4}=-\frac{3}{2}\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=3\\ x=-1\end{matrix}\right.\)
Bài 3:
a) Ta thấy:
\(|x+\frac{15}{19}|\geq 0, \forall x\Rightarrow A\ge 0-1=-1\)
Vậy GTNN của $A$ là $-1$ khi \(x+\frac{15}{19}=0\Leftrightarrow x=-\frac{15}{19}\)
b)Vì \(|x-\frac{4}{7}|\geq 0, \forall x\Rightarrow B\geq \frac{1}{2}+0=\frac{1}{2}\)
Vậy GTNN của $B$ là $\frac{1}{2}$ khi \(x-\frac{4}{7}=0\Leftrightarrow x=\frac{4}{7}\)
\(a,C=\left|\dfrac{1}{3}x+4\right|+1\dfrac{2}{3}\)
Ta có \(\left|\dfrac{1}{3}x+4\right|\ge0\)
\(\Rightarrow\left|\dfrac{1}{3}x+4\right|+1\dfrac{2}{3}\ge1\dfrac{2}{3}\)
Dấu "=" xảy ra khi \(\left|\dfrac{1}{3}x+4\right|=0\)
\(\Leftrightarrow\dfrac{1}{3}x+4=0\)
\(\Leftrightarrow\dfrac{1}{3}x=0-4=-4\)
\(\Leftrightarrow x=-4:\dfrac{1}{3}\)
\(\Leftrightarrow x=-12\)
Vậy \(\min\limits_C=1\dfrac{2}{3}\Leftrightarrow x=-12\)
\(b,D=\left|x-6\right|+\left|x+\dfrac{5}{4}\right|\)
Ta có : \(\left\{{}\begin{matrix}\left|x-6\right|\ge-x+6\\\left|x+\dfrac{5}{4}\right|\ge x+\dfrac{5}{4}\end{matrix}\right.\)
\(\Rightarrow\left|x-6\right|+\left|x+\dfrac{5}{4}\right|\ge-x+6+x+\dfrac{5}{4}=\dfrac{29}{4}\)
Dấu "=" xảy ra khi
\(\left\{{}\begin{matrix}-x+6\ge0\\x+\dfrac{5}{4}\ge0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\le6\\x\ge-\dfrac{5}{4}\end{matrix}\right.\)
Vậy \(\min\limits_D=\dfrac{29}{4}\Leftrightarrow-\dfrac{5}{4}\le x\le6\)
b) \(D=\left|x-6\right|+\left|x+\dfrac{5}{4}\right|\)
\(D=\left|6-x\right|+\left|x+\dfrac{5}{4}\right|\ge\left|6-x+x+\dfrac{5}{4}\right|=\dfrac{29}{4}\)
Dấu = xảy ra khi \(\left(6-x\right)\left(x+\dfrac{5}{4}\right)\ge0\Leftrightarrow-\dfrac{5}{4}\le x\le6\)
vậy \(D_{min}=\dfrac{29}{4}\) khi \(-\dfrac{5}{4}\le x\le6\)
Giải:
a) Có: \(A=\left|x-\dfrac{3}{4}\right|+1\)
Vì \(\left|x-\dfrac{3}{4}\right|\ge0\forall x\)
\(\Leftrightarrow\left|x-\dfrac{3}{4}\right|+1\ge1\forall x\)
Hay \(A\ge1\forall x\)
Vậy giá trị nhỏ nhất của A là 1.
b) \(B=5-\left|\dfrac{2}{3}-x\right|\)
Vì \(\left|\dfrac{2}{3}-x\right|\ge0\forall x\)
\(5-\left|\dfrac{2}{3}-x\right|\le5\)
Hay \(B\le5\forall x\)
Vậy giá trị lớn nhất của B là 5.
Chúc bạn học tốt!!!
Đặt:
\(HIEUCANCER=\left|x-\dfrac{3}{4}\right|+1\)
\(\left|x-\dfrac{3}{4}\right|\ge0\forall x\in R\)
\(HIEUCANCER=\left|x-\dfrac{3}{4}\right|+1\ge1\)
Dấu "=" xảy ra khi:
\(x=\dfrac{3}{4}\)
Đặt:
\(HIEUBD=5-\left|\dfrac{2}{3}-x\right|\)
\(\left|\dfrac{2}{3}-x\right|\ge0\forall x\in R\)
\(HIEUBD=5-\left|\dfrac{2}{3}-x\right|\le5\)
Dấu "=" xảy ra khi:
\(x=\dfrac{2}{3}\)
\(P=\left(0,5-\dfrac{3}{5}\right):\left(-3\right)+\dfrac{1}{3}-\left(-\dfrac{1}{6}\right):\left(-2\right)\)
\(=\left(-\dfrac{1}{2}-\dfrac{3}{5}\right):\left(-3\right)+\dfrac{1}{3}-\left(-\dfrac{1}{6}\right).\left(-\dfrac{1}{2}\right)\)
\(=\left(\dfrac{-5-6}{10}\right):\left(-3\right)+\dfrac{1}{3}-\dfrac{1}{12}\)
\(=-\dfrac{11}{10}:\left(-3\right)+\dfrac{1}{4}\)
\(=-\dfrac{11}{10}.\left(-\dfrac{1}{3}\right)+\dfrac{1}{4}=\dfrac{11}{30}+\dfrac{1}{4}=\dfrac{37}{60}\)
Vậy \(P=\dfrac{37}{60}\)
\(Q=\left(\dfrac{2}{25}-1,008\right):\dfrac{4}{7}:\left[\left(3\dfrac{1}{4}-6\dfrac{5}{9}\right):2\dfrac{2}{17}\right]\)
\(=\left(\dfrac{2}{25}-\dfrac{126}{125}\right):\dfrac{4}{7}:\left[\left(\dfrac{13}{4}-\dfrac{59}{9}\right).\dfrac{36}{17}\right]\)
\(=-\dfrac{116}{125}.\dfrac{7}{4}:\left(-\dfrac{119}{36}.\dfrac{36}{17}\right)\)
\(=\dfrac{-29.7}{125}:\left(-7\right)=\dfrac{29}{125}\)
Vậy \(Q=\dfrac{29}{125}\)
Sau khi thực hiện phép tính ta được kết quả các giá trị:
\(A=\dfrac{1}{3}\) \(B=-5\dfrac{5}{12}\) \(C=-0,22\)
Sắp xếp: \(-5\dfrac{5}{12}< -0,22< \dfrac{1}{3}\) tức là \(B< C< A\)
Khi tính xong giá trị biểu thức A , B và C ta được kết quả như sau :
\(A=\dfrac{1}{3}\) ; \(B=-5\dfrac{5}{12}\); \(C=-0,22\)
Sắp xếp : \(B< C< A\)\(\left(-5\dfrac{5}{12}< -0,22< \dfrac{1}{3}\right)\)
a: \(A=\left|x-\dfrac{7}{4}\right|+\dfrac{8}{5}>=\dfrac{8}{5}\)
Dấu = xảy ra khi x=7/4
b: \(B=\left|5-x\right|+\left|x+\dfrac{3}{4}\right|>=\left|5-x+x+\dfrac{3}{4}\right|=\dfrac{23}{4}\)
Dấu = xảy ra khi (x-5)(x+3/4)<=0
=>-3/4<=x<=5